Triangle Inequalities: An interesting tool(Feb 24, 2009)

We know that proving triangular inequalities are interesting and somewhat tough, but I use a very useful theorem that is handy in problem solving.

Let ABC be a triangle with sides a, b, c. Let I be its incentre and let the incircle touch the sides AB, BC, CA on D, E, F respectively. Then we have (from showing congruences): AD = AF, BD = BE, CF = CE. Refer to the image.
So, we have that a = x + y, b = y + z, c = z + x for some x, y, z.
Image
PROOF
Refer to the image.
Join DI, EI, FI, AI, BI, CI.
Now, in \Delta ADI and \Delta AFI
\angle AFI = ADI = 90\degree
AI is common.
DI = FI
Hence, from the RHS rule of congruency these two are congruent.
\Delta ADI and \Delta AFI are congruent , giving AD = AF
Similarly we have BD = BE & CF = CE
QED.

APPLICATIONS
1. In a triangle ABC the bisectors AD, BE, CF meet at I. Prove that:
\frac {1}{4} < \frac {IA}{AD}\frac {IB}{BE}\frac {IC}{CF}\le \frac {8}{27}
(IMO 1991/1)

2. Prove that, if a, b, c are the sides of a triangle, we have:
\frac {a}{b + c - a} + \frac {b}{c + a - b} + \frac {c}{a + b - c}\ge 3
(RMO, India)

3. For any triangle with sides a, b, c, Prove that we have
a^2b(a - b) + b^2c(b - c) + c^2a(c - a)\ge 0
(proposed by Klamkin, IMO 1983/6)

Solutions to these problems are welcome.
Thanking you.

Incircle.JPG
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